Stabilization of the wave equation by on-off and positive-negative feedbacks

URL:http://www.emath.fr/cocv/ DOI: 10.1051/cocv:2002015

STABILIZATION OF THE WAVE EQUATION BY ON-OFF AND POSITIVE-NEGATIVE FEEDBACKS^∗

Patrick Martinez

and Judith Vancostenoble

Abstract. Motivated by several works on the stabilization of the oscillator by on-off feedbacks, we study the related problem for the one-dimensional wave equation, damped by an on-off feedbacka(t)ut. We obtain results that are radically different from those known in the case of the oscillator. We consider periodic functions a: typicallyais equal to 1 on (0, T), equal to 0 on (T, qT) and isqT-periodic. We study the boundary case and next the locally distributed case, and we giveoptimal results of stability.

In both cases, we prove that there are explicit exceptional values of T for which the energy of some solutions remains constant with time. IfT is different from those exceptional values, the energy of all solutions decays exponentially to zero. This number of exceptional values iscountablein the boundary case and finitein the distributed case. When the feedback is acting on the boundary, we also study the case ofpostive-negative feedbacks: a(t) =a0>0 on (0, T), anda(t) =−b0<0 on (T, qT), and we give the necessary and sufficient condition under which the energy (that is no more nonincreasing with time) goes to zero or goes to infinity. The proofs of these results are based on congruence properties and on a theorem of Weyl in the boundary case, and on new observability inequalities for the undamped wave equation, weakening the usual “optimal time condition” in the locally distributed case. These new inequalities provide also new exact controllability results.

Mathematics Subject Classification. 35L05, 35B35, 35B40, 11A07.

Received June 25, 2001. Revised November 27, 2001.

Contents

1. Introduction 336

2. Main results 338

2.1. Main results for boundary on-off dampings 338

2.2. Main results for boundary positive-negative feedbacks 339

2.3. Main results for locally distributed on-off feedbacks 340

2.4. “On-off” observability inequalities 340

3. Relation to literature 341

3.1. The time-independent case 341

3.2. The time-dependent case 342

3.3. The on-off case 342

Keywords and phrases:Damped wave equation, asymptotic behavior, on-off feedback, congruences, observability inequalities.

∗This work was partially made when the first author was working in the ENS Cachan, Antenne de Bretagne.

1M.I.P. Universit´e Paul Sabatier Toulouse III, 118 route de Narbonne, 31062 Toulouse Cedex 4, France;

e-mail:martinez@mip.ups-tlse.fr, vancoste@mip.ups-tlse.fr

c EDP Sciences, SMAI 2002

4. Comments on the results and optic rays propagation 343

4.1. Comments in the case of boundary feedback 343

4.2. Comments in the case of locally distributed feedback 344

5. Other results and comments on the proofs 344

5.1. Other results in the case of a boundary on-off feedback 344

5.2. Remarks on the proof in the case of a locally distributed feedback 346 6. Proof of Proposition 5.1 (on-off boundary feedback whenq= 2) 350

6.1. The useful formula for the energy 350

6.2. Proof of Proposition 5.1 351

7. Proof of Theorem 2.1 (on-off boundary feedback in the general case) 353

7.1. Stability if T is different of some exceptional values 353

7.2. Non stability if T is equal to one of those exceptional values 353

8. Proof of Theorem 2.2 (positive-negative boundary feedbacks) 354

8.1. The cyclic case: 1/T ∈Q\N 355

8.2. The equidistributed case: 1/T 6∈Q 356

9. Proof of Proposition 5.2 (estimate of the extinction time) 358

10. Proof of Theorem 2.3 when q= 2 (locally distributed on-off feedback) 361

10.1. The useful tool 361

10.2. Proof of Theorem 2.3 when q = 2 andω= (0,1) 362

10.3. Proof of Theorem 2.3 when q = 2 andω= (1/2−λ,1/2 +λ)in the case1/T 6∈2N 363 10.4. Proof of Theorem 2.3 when q = 2 and ω = (1/2−λ,1/2 +λ)in the case 1/T ∈ 2N and

2λ < T 365

10.5. Proof of Theorem 2.3 when q = 2 andω= (1/2−λ,1/2 +λ)and1/T ∈2Nand2λ > T 366

11. Proof of Theorem 5.2 (observability inequalities) 366

11.1. Proof of Lemma 11.1 368

11.2. Proof of Lemma 11.2 371

11.3. Proof of Lemma 11.3 371

11.4. Proof of Lemma 11.4 373

11.5. Proof of Lemma 11.5 373

12. Tools for the proof of Theorem 2.3 and Theorem 2.4 373

13. Proof of Proposition 10.1 (link between stabilization and observability) 374

13.1. Equation (10.3) implies (10.4) 374

13.2. Equation (10.4) implies (10.3) 375

References 376

1. Introduction

Motivated by several works on ordinary differential equations [1, 11, 12, 31, 34, 35], we consider first the wave equation in one space dimension, damped by aboundary on-off feedback a(t)ut, where a : R⁺ −→ R⁺ is a bounded nonnegative function that can sometimes be equal to zero:











utt−uxx= 0, x∈(0,1), t≥0,

u(0, t) = 0, t≥0,

ux(1, t) =−a(t)ut(1, t), t≥0, (u(x,0), ut(x,0)) = (u⁰(x), u¹(x)), x∈(0,1),

(1.1)

where (u⁰, u¹) is given inV ×L²(0,1) (withV ={v∈H¹(0,1) | v(0) = 0}).

Next we consider the wave equation in one space dimension, damped by alocally distributed on-off feedback a(t)χω(x)ut, whereω⊂(0,1):







utt−uxx=−a(t)χω(x)ut, x∈(0,1), t≥0, u(0, t) =u(1, t) = 0, t≥0,

(u(x,0), ut(x,0)) = (u⁰(x), u¹(x)), x∈(0,1),

(1.2)

where (u⁰, u¹) is given inH₀¹(0,1)×L²(0,1).

In both cases we define the energy ofuby

∀t≥0, Eu(t) = 1 2

Z 1 0

(ux2

(x, t) +ut2

(x, t)) dx.

We are interested in the asymptotic behavior of the energy. We consider the typical case where

a(t) =a0>0 fort∈[0, T), a(t) = 0 fort∈[T, qT) andaisqT-periodic (1.3) for some integerq≥2. In both cases, we prove that there is aset of explicit exceptional values forT for which the energy of some solutions remains constant with time. IfT is not one of those exceptional values, the energy decays exponentially to zero (and the moreT is close to such an exceptional value, the slowlier it decays). This set iscountablein the boundary case, given by

1 T ∈

q[−1 p=1

q 2pN,

and onlyfinite in the locally distributed case: ifω= ((1/2)−λ,(1/2) +λ)⊂(0,1), it is given by 1

T ∈

q[−1 p=1

q

pN and (q−1)T >2λ.

We also characterize (in term of support) the initial conditions that create solutions whose energy does not decay to zero.

Motivated by a question of Zuazua and [7], we consider alsopositive-negative feedbacks for (1.1): a isqT- periodic and

a(t) =a0>0 for t∈[0, T), a(t) =−b0<0 for t∈[T, qT). (1.4) Note that in this case the energy is only nonincreasing during the time intervals [mqT, mqT +T), and non- decreasing during the other time intervals. IfT is exceptional, then the energy of some solutions increases to infinity. IfT is not exceptional, we give the necessary and sufficient condition that tells that, roughly speaking, the energy of all solutions goes exponentially to zero, or the energy of some solutions go exponentially to infinity:

for example if 1/T /∈Q, then denoting K0:=

a0−1 a0+ 1

1/q b0+ 1 b0−1

(q−1)/q

;

stability holds if and only ifK0<1. In particular, note that stability always holds ifa0= 1 (for allb06= 1).

For the boundary case, our proofs are based on congruence properties and a theorem of Weyl. For the distributed case, they are based on new observability inequalities: “on-off” observability inequalities, that tells

that, roughly speaking, ifT0 is the minimal time that one need to observe the solution, then observing it “half of the time” can be sufficient (in fact only some parts of the time interval (0, T0) are useful).

Our results areradically differentfrom those of the wave equation damped by linear time-dependent feedbacks a(t)u⁰ when a is always positive (possibly decaying to zero), and even radically different from those of the ordinary differential equations damped by on-off feedbacks.

More precisely, our results are the following:

2. Main results

First consider the problem (1.1) and the following functiona:

a(t)

0 T qT

t

Figure 1.

Such on-off feedback laws have been widely studied in the case ofordinarydifferential equations, but seem to have never been studied forpartialdifferential equations. We prove the following:

Theorem 2.1. Assume (1.3). For all (u⁰, u¹) ∈ V ×L²(0,1), there exists a unique u solution of (1.1).

Moreover, (i) if

1 T ∈

q[−1 p=1

q

2pN, (2.1)

there exists some (u⁰, u¹)∈V ×L²(0,1) such thatEu(t) remains constant with time: Eu(t) =Eu(0) >0for allt≥0;

(ii) if

1 T ∈/

q[−1 p=1

q

2pN, (2.2)

then for all (u⁰, u¹)∈V ×L²(0,1), the energy Eu(t) of the solutions of (1.1) decays uniformly exponentially to0(or achieves zero in finite time in the particular case a0= 1).

Remarks.

1. The non stability result in the exceptional case (2.1) isradically differentfrom what happens for ordinary differential equations, or even for partial differential equations like (1.1) when the functionadecreases to zero remaining always positive (see [25]).

2. We alsocharacterize(in term of support) the initial conditions that create solutions whose energy does not decay to zero. Moreover, in the non-exceptional case (2.2), we provide estimates on the “extinction time” (where the energy is equal to zero), and we give optimal estimates whenq= 2 andq= 3.

3. We also give explicit and optimal estimate of the exponent of the exponential decay of the energy or of the extinction time in the particular casea0= 1. See Section 5.

2.2. Main results for boundary positive-negative feedbacks

Now we consider the more general case ofpositive-negative feedbacks: the functionais now 2T-periodic to simplify:

T 2T

a(t)

t

Figure 2.

In this case, note that the energy is only nonincreasing on the time intervals [2mT,2mT+T) and nondecreas- ing the other time intervals. There are some works when the feedback is of the typeb(x)u⁰, where the function bdepends onx(and not ont) and is of indefinite sign, but “more positive than negative” (see,e.g., Freitas and Zuazua [7], Benaddi and Rao [5]) But to our knowledge, such time dependent positive-negative feedbacks laws have never being studied. We prove the following:

Theorem 2.2. Assume (1.4)with q=2. Assume thatb06= 1. Then for all(u⁰, u¹)∈V ×L²(0,1), there exists a unique usolution of (1.1). Moreover,

(i) if T satisfies (2.1), that is simply 1/T ∈ N in this case (q = 2), then there exists some (u⁰, u¹)

∈V ×L²(0,1)such that Eu(t)goes exponentially to infinity ast→ ∞; (ii) if 1/T =p⁰/q⁰ wherep⁰ andq⁰ are relatively primes, denote







KT :=

a₀−1 a0+1

1/2

b₀+1 b0−1

1/2

if q⁰ is even, KT :=

a₀−1 a0+1

(q⁰−1)/2q⁰

b₀+1 b0−1

1−(q⁰−1)/2q⁰

ifq⁰ is odd;

(2.3)

then ifKT <1, the energy of all solutions goes exponentially to zero (as K_T^t), an if KT >1, the energy of some solutions goes exponentially to infinity (asK_T^t);

(iii) if1/T /∈Q, denote

K0:=

a0−1 a0+ 1

1/2 b0+ 1 b0−1

1/2

; (2.4)

then ifK0<1, the energy of all solutions goes exponentially to zero (as K₀^t), an if K0>1, the energy of some solutions goes exponentially to infinity (asK₀^t).

Remarks.

1. It is interesting to note that if a0 = 1, then for all values of b0 (except b0 = 1 for which we have no solution), we obtain exponential decay.

2. Part (iii) relies on a theorem of Weyl: if θ /∈ Q, then the sequence ({nθ})n is not only dense but also equidistributed in [0,1). (As usual,{x}denotes the fractional part ofx.) Consequently, the critical value K0 that appears in (iii) does not depend onT.

3. Theorem 2.2 could easily be extended to the general case q ≥ 2; the critical value in part (iii) would become

K˜0:=

a0−1 a0+ 1

1/q b0+ 1 b0−1

1−1/q

·

2.3. Main results for locally distributed on-off feedbacks

Now we consider the problem (1.2), whereω is the open nonempty subset ((1/2)−λ,(1/2) +λ) of (0,1),a is the time periodic function (1.3) and (u⁰, u¹) is given inH0¹×L²(0,1). We prove the following:

Theorem 2.3. Assume (1.3)and assume that 0< λ≤1/2.

(i)If

1 T ∈

q[−1 p=1

q

pN and (q−1)T >2λ,

then there exists initial conditions(u⁰, u¹)∈H₀¹×L²(0,1)such that the energy of the solutions of (1.2)remains constant with time: Eu(t) =Eu(0)>0for all t≥0.

(ii)If

1 T ∈

q[−1 p=1

q

pN and (q−1)T <2λ

!

, or 1

T 6∈

q[−1 p=1

q pN

!

, (2.5)

then the energy of the solutions of (1.2)decays uniformly exponentially to0.

Once again, this isradically differentfrom what happens for ordinary differential equations, or even for the damped wave equation when the functionadecreases to zero remaining always positive. In the case (2.5), we prove the uniform decay of the energy thanks tonew observability inequalities:

2.4. “On-off” observability inequalities Considering the undamped problem







φtt−φxx= 0, x∈0,1), t≥0, φ(0, t) =φ(1, t) = 0, t≥0,

(φ(x,0), φt(x,0)) = (φ⁰(x), φ¹(x)), x∈(0,1),

(2.6)

it is well known that if 0< a < b <1 andT^∗>2 max(a,1−b), then the solutions of (2.6) satisfy the following observability inequality

Eφ(0) ≤ C Z T^∗

0

Z b a

φ²_t(x, t) dxdt (2.7)

for some positive constantC=C(T^∗) (see Haraux [9] and Zuazua [37]). This is optimal in the sense that you cannot have this inequality with someT^∗ <2 max(a,1−b). In our case, for example, ifλ >1/8, then we can apply it withT^∗= 3/4 and we obtain

Eφ(0) ≤ C Z 3/4

0

Z (1/2)+λ (1/2)−λ

φ²t(x, t) dxdt. (2.8)

We improve this inequality showing that Eφ(0) ≤ C

Z 1/4 0

Z (1/2)+λ (1/2)−λ

φ²t(x, t) dxdt+C Z 3/4

1/2

Z (1/2)+λ (1/2)−λ

φ²t(x, t) dxdt. (2.9) This is coherent with the fact that each optic ray touches the damping region during the time intervals (0,1/4) or (1/2,3/4).

More generally, we prove the following:

Theorem 2.4. Assume 1/T ∈ Sq−1 p=1 q

pN and (q−1)T < 2λ. Then there exists C > 0 such that, for all φ solution of (2.6),

Eφ(0) ≤C Z q−1

0

a(t) Z

ω

φ²_t(x, t) dxdt. (2.10)

Note that this obviously also givesnew exact controllability results, applying the method H.U.M. of J.-L. Lions [22]: given (u⁰, u¹)∈H₀¹(Ω)×L²(Ω), there exists a controlh∈L²((0,1/4)×(1/2,3/4), L²(ω)) such that the solution of the problem







utt−uxx=χω(x)h(t, x), x∈(0,1), t≥0, u(0, t) =u(1, t) = 0, t≥0,

(u(x,0), ut(x,0)) = (u⁰(x), u¹(x)), x∈(0,1),

(2.11)

satisfiesu(3/4) = 0 =ut(3/4). This implies that the controlh that drives the system from the state (u⁰, u¹) to the resthas only to act on the time intervals(0,1/4) and(1/2,3/4). Our observability inequalities or exact controllability results are coherent with the general results related on the rays propagation for systems with time independent coefficients (see [2, 3]). Moreover we provide a direct proof of these inequalities, and we would like to emphasize that our method can also provide results for semilinear wave equations (see [26]).

3. Relation to literature

We are interested in the asymptotic behavior of the energy of the solutions of (1.1). First we recall that if the functionaisconstant:

a(t) =a0>0 for all t∈R⁺,

then for all (u⁰, u¹) given inV ×L²(0,1), there exists a unique solutionuof (1.1) and its energyEu(t) decays exponentially to 0 astgoes to infinity. More precisely, it is easy to prove that:

– ifa0= 1, thenEu(t) achieves 0 infinite timein time t= 2: Eu(2) = 0;

– ifa06= 1 thenEu(t)→0 exponentially ast→ ∞with an explicit decay rate that depends ona0:

∀t≥0, Eu(t)≤Eu(0)e^{−ω(t/2−1)} withω= 2 lna0+ 1 a0−1

>0.

On the other hand, with the same feedbacka(t) = a0>0 for all t∈R⁺, exponential decay of the solutions of (1.2) follows easily from the multiplier method.

Of course, in both cases, the exponential decay of the energy for this problem is also an easy consequence of the general “optic rays condition” of Bardoset al.[2]: it is clear that each optic ray touches the damping region (the point 1) in time at most 2 in the case of boundary damping, and crosses the regionω in time at most 2 in the case of locally distributed damping.

Remark. The proof of theextinction in finite timefollows directly from D’Alembert Formula (see the beginning of the proofs of Th. 2.1, (ii) or Prop. 5.1). Note that extinction in finite time was proved in [16] (see Th. 0.5, p. 6): for the wave equation with a boundary feedback acting atboth extremities (x= 0 andx= 1), Komornik obtained the extinction in time t = 1. In our case, since the feedback only acts at oneextremity (x= 1), we have the same phenomenon but we need twice more time before the extinction.

Note that this phenomenon is related to theradiation boundary conditions(see [4]), which correspond to the fact that there is no reflection from the boundary into the domain. Indeed, in the case of the wave equation, the boundary conditionux(1, t) +ut(1, t) = 0 is a radiation boundary condition.

3.2. The time-dependent case

Now we consider the asymptotic behavior of the energy undertime-dependentfeedback laws. This problem has been largely studied when the damping term is “not too small” (in order to preventunderdamping) or “not too large” (to preventoverdamping), that means when there exists somepositiveand nonincreasing continuous functionσ:R⁺−→R⁺that satisfies

∀t≥0, σ(t)≤a(t)≤ 1

σ(t), (3.1)

Z +∞ 0

σ(τ) dτ=∞. (3.2)

In this case, for both problem (1.1) and problem (1.2), the energy decays with an explicit decay rate: there existsω >0 such that

∀t≥0, Eu(t)≤Eu(0)e^1−ω

Rt 0σ(τ) dτ

. (3.3)

On the other hand, if R+∞

0 a(τ) dτ < ∞, there are some solutions whose energy does not decay to 0. (For more general results about stabilization properties using nonlinear time-dependent dampings g(t, ut), see, e.g. [25, 27, 32] and the references therein.)

3.3. The on-off case

Here we are interested in the asymptotic behavior of the energy underon-off dampings: we assume that the function amay vanish on non-zero measure sets, and we want to study the asymptotic behavior of the energy.

To our knowledge, such works have not yet been done in the case of partial differential equations. Bardos et al.[2] (p. 1029) noted just thatinvisiblesolutions may appear in the case of time-dependent coefficients, and their existence has to be studied on each case. Note also that in our case the family of invisible solutions is not at all finite dimensional.

However this has been widely studied in the case ofordinarydifferential equations (see [1, 11, 12, 31, 34, 35]).

The typical problem is the oscillator damped by an on-off damping

u⁰⁰+a(t)u⁰+u= 0; (3.4)

the energy decays to zero if the damping is “sufficiently active”: in the case where the functionais equal to 1 on a sequence of disjoint intervalsIn and nonnegative elsewhere, then the energy decays to 0 if

X

n

|In|³=∞, (3.5)

and 3 is the best possible exponent (see [31]). In particular, this implies that under (3.5), the localization of the damping time intervals In is not important; Pucci and Serrin noted that “the exact switching times can be of great importance” (p. 831 in [31]) in some situations where (3.5) is not satisfied but 0 is still a global attractor of the problem if the switching times are correctly set. More general cases are studied, in particular the case where the functionais equal to some positive constantmn onIn, and then 0 is a global attractor of the problem if some condition like (3.5) is satisfied (the condition lies on the divergence of some series whose general term contains|In|and min (mn,_m¹_n) in order to prevent underdamping and overdamping).

The case of partial differential equations damped by on-off dampings (applied at the boundary or locally distributed in the domain) isradically different from the case of ordinary differential equations: even in the simplest case where the functionais equal to 1 on [0, T) and to 0 on [T,2T), and is 2T-periodic, there are values ofT for which the energy does not decay to zero, although the conditionsR+∞

0 a(τ) dτ=∞andP

n|In|³=∞ (whereIn := (2nT,(2n+ 1)T) are clearly satisfied.

4. Comments on the results and optic rays propagation

Our proof is based on D’Alembert formula and oncongruence properties. In fact this congruence properties are equivalent to study the optic rays propagation; we prove that ifT is not one of those exceptional values, each ray touches the boundary pointx= 1 (where the dissipative condition is applied) in time at most 2NT+ 2, where NT depends onT and at an instant time t where the damping is effective. In Figure 3 we present an example where the value ofT is exceptional (T = 1/5). We can see that there exists a ray that touches the boundary only at times when the feedback is non active (or of the wrong sign in the case of positive-negative feedbacks):

0 1 2 3 4

1 x

t

Figure 3.

In Figure 4, we present an example where the value ofT is not exceptional (T = 2/5). In this case, we can see that each ray will touch the damping region at a time when the feedback is active.

This property is quite natural with respect to the general optic rays condition for time independent feed- backs [2]. However, we do not know if this theory can be adapted in this case of a time dependent feedback a(t)ut(1, t) (and moreover with a function a that is not continuous), and even if it is true, it is not easy to determine explicitly the exceptional values ofT.

0 1 2 3 4 1

x

t

Figure 4.

4.2. Comments in the case of locally distributed feedback

We are interested in the asymptotic behavior of the energyEu of the solutions of (1.2). First we recall that if the functionaisconstantonR⁺, thenEudecays exponentially to 0, for all nonempty open subsetωwherever its location in (0,1) (see,e.g.[8]). A geometrical explanation of this is that each optic ray touches the damping regionωin time at most 2 (see [3]). (Note that in the case of a symmetrical open subsetω:= (1/2−λ,1/2 +λ) with 0< λ≤1/2, each optic ray touches the damping region in time at most 1).

Here we consider the case whereais the periodic function defined by (1.3) and whereω := (1/2−λ,1/2 +λ) with 0< λ≤1/2 (λ= 1/2 means that the feedback is uniformly distributed in (0,1)). There is no problem for the existence and the regularity of the solution, solving successively on the time intervals (0, T), (T,2T), ...

First consider the case of auniformly distributeddamping,i.e.we assumeλ= 1/2. In this case, note that it is clear that each optic ray crosses the damping region during a period when the damping is effective. In this case, we prove positive results of exponential stabilization.

Next consider the more interesting case of alocally distributeddamping: 0< λ <1/2. Note carefully that there are some values ofT and some values ofλfor whichsome rays cross the damping region when the feedback is non active. For example, take T = 1/2, λ < T /2 = 1/4, and consider the optic rays that leaves the point x = T /2 = 1/4 and that goes to the left (towards the point x = 0) at time t = 0: this ray describes the segment [1/4,3/4] (that contains the dissipative region) in direct sens or in the other sens during the time intervals [T,2T], [3T,4T], ..., thus during periods whena(t) = 0. The same situation occurs if 1/T ∈2Nwith 2λ < T. We obtain negative results of exponential stabilization in all these cases, and positive results in the other cases, which is coherent with the optic rays condition known for time independent feedbacks [2].

Note that the situation is muchmore complex than the case of boundary damping: even when T takes some

“exceptional” values, we can still have exponential decay of the energy of the solutions, provided that the damping region is large enough. Note also that when the damping region is “large enough” (in particular when ω= (0,1), then we find a result analogous to the one related toordinary differential equation, since we obtain stabilization for allT >0.

At last, note that the previouspositive results of uniform exponential stabilization are still true if we just assume that a ∈ L^∞(R⁺;R⁺) is qT-periodic such that a(t) ≥ a0 > 0 for t ∈ [0, T). On the same way, the negative results of uniform exponential stabilization are still true if we just assume that a ∈ L^∞(R⁺;R⁺) is qT-periodic such thata(t) = 0 fort∈[T, qT).

5. Other results and comments on the proofs

5.1.1. Optimal estimates of the extinction time

We complete Theorem 2.1 giving optimal estimates of the “extinction time”τT = 2NT+ 2, where the energy achieves zero in the particular casea0= 1. Note that this also give optimal estimate of the exponential energy

decay in the general case since it is given by

∀t≥0, Eu(t)≤Eu(0)e^{−ω(t/(2NT+2)−1)}.

We denote by d(x,N) the distance betweenxandN. We assume that T <2 since if T ≥2, then we already know thatEu(t) = 0 for allt≥2.

First consider the typical case q = 2. We recall from Theorem 2.1 that if 1/T ∈ N, the energy of some solutions does not decay to zero. In the other case, we give optimal estimates on the “extinction time”:

Proposition 5.1. Assume (1.3) with q = 2. Then if 1/T 6∈ N, then for all (u⁰, u¹) ∈ V ×L²(0,1), Eu(t) achives zero in finite time. More precisely,

∀t≥2NT + 2, Eu(t) = 0, (5.1)

whereNT is the smallest integer such that

2NTd 1

T,N

≥1. (5.2)

Moreover, NT is optimal in the following sense: there exists some (u⁰, u¹)∈ V ×L²(0,1) such that Eu(2NT)

=Eu(0)>0.

Next we look to the general case. We also give optimal estimates of the “extinction time” whenq= 3 and we give general estimates of the “extinction time” in the general case:

Proposition 5.2. Assume (1.3)and assume (2.2).

(i)Assume thatq= 3. If

d 2

3T,N

≤ 1

3, (5.3)

letNT be the smallest integer such that

3d 2

3T,N

NT ≥2;

and if

1 3< d

2 3T,N

<1

2, (5.4)

let N_T⁰ be the smallest integer such that

1−2d 2

3T,N

NT⁰ ≥2 3−d

2 3T,N

, (5.5)

andNT := 2N_T⁰ −1; thenEu(t) = 0for all t≥2NT+ 2. Moreover,NT is optimal in the following sense: there exists some initial conditions such thatEu(2NT) =Eu(0)>0.

(ii)In the general case, we have the following estimate: Eu(t) = 0 for all t ≥2mN+ 2, where m and N are integers chosen such that

0<

m

2 qT

≤1

q and N

m 2

qT

≥q−1

q (5.6)

(where{x}denotes the fractional part ofx).

Remark. Such malways exists under (2.2). It would be interesting to find in a general way the best time τT

for whichEu(t) = 0 ift≥τT, like we did forq= 2 andq= 3.

5.1.2. Extension to nonlinear feedbacks

Combining the methods used in [36] with the method used for the proof of Proposition 5.1, we can also study the wave equation damped by a boundarynonlinear on-off feedback:











utt−uxx= 0, x∈(0,1), t≥0,

u(0, t) = 0, t≥0,

ux(1, t) =−a(t)g(ut(1, t)), t≥0, (u(x,0), ut(x,0)) = (u⁰(x), u¹(x)), x∈(0,1),

where ais defined by (1.3) and wheregis increasing. Once again, whenT is exceptional, we have not stability, and whenT is not exceptional, we easily prove that there is still uniform stabilization, with explicit decay rate that depends ong.

5.2. Remarks on the proof in the case of a locally distributed feedback 5.2.1. Link between on-off stabilization and on-off observability

To prove Theorem 2.3 in the typical caseq= 2, we firstreduce the problem to the proof of an observability inequalityfor the wave equation without damping (see Prop. 10.1): there is uniform exponential stabilization of solutions of (1.2) if and only if the solutions of (2.6) satisfy, for someT^∗>0,

Eφ(0) ≤ C Z T^∗

0

a(t) Z

ω

φ²_t(x, t) dxdt. (5.7)

Then to prove this inequality, we will have to distinguish the following different cases:

• λ= 1/2 andT >0. It is the case of a uniformly distributed damping. In this case, we can easily prove the observability inequality using spectral decomposition of the solutions of the wave equation;

• λ <1/2 and 1/T /∈2N. In this case, we use the fact thatT is not an exceptional value as we did in the case of boundary feedback, to obtain similar congruence properties. And combining these properties with the spectral decomposition made in the previous case, we deduce the observability inequality;

• λ <1/2, 1/T ∈2Nand 2λ < T. In this case, we construct solutions that never “see the damping” such that the observability inequality does not hold;

• λ <1/2, 1/T ∈2Nand 2λ > T. To treat this case, we need new observability inequalities (see Prop. 5.2) presented in the following section.

5.2.2. On-off observability inequalities

Consider the problem (2.6). The following result is well known:

Theorem 5.1(Haraux [9]). Let (a, b)⊂(0,1) be a given interval. Set T₀^∗= 2max (a,1−b).

Then for allT^∗> T₀^∗ , there existsC >0such that, for allφ solution of (2.6), Eφ(0) ≤ C

Z T^∗ 0

Z b a

φ²_t(x, t) dxdt. (5.8)

This is optimal in the sense that you cannot have this inequality with someT^∗<2 max(a,1−b). In our case, the condition “T^∗>2 max(a,1−b)” becomes “T^∗+ 2λ >1”. Consequently,if we assume 2λ >1−T, we can directly deduce uniform exponential stabilization from Theorem 5.1.

For example, consider the case q = 2 and T = 1/2. Note that in this case, the condition “2λ > T” of Theorem 2.3 is exactly the same as the previous condition “2λ > 1−T” i.e. it is “λ > 1/4”. So we can apply (5.8) withT^∗=T = 1/2 and we obtain

Eφ(0) ≤ C Z 1/2

0

Z (1/2)+λ (1/2)−λ

φ²_t(x, t) dxdt=C Z 1

0

a(t) Z

ω

φ²_t(x, t) dxdt, (5.9)

which is exactly what we need to prove that the energy decays exponentially to zero in this case. This proves Theorem 2.3 in the caseq= 2 andT = 1/2.

For T 6= 1/2, the two conditions are not the same. We still can deduce results of uniform exponential stabilization from Theorem 5.1, but it is not sufficient to prove Theorem 2.3.

For example, consider the case T = 1/4. If we assume λ > (1−T)/2 = 3/8, we can apply (5.8) with T^∗=T = 1/4 and we obtain

Eφ(0) ≤ C Z 1/4

0

Z (1/2)+λ (1/2)−λ

φ²_t(x, t) dxdt≤C Z 1

0

a(t) Z

ω

φ²_t(x, t) dxdt, (5.10)

which proves that the energy decays exponentially to zero in this case. Note that the assumption “λ >(1− T)/2 = 3/8” and this inequality correspond, as we can see in Figure 5, to the fact that each optic ray crosses the damping region during the period (0, T) = (0,1/4).

x

0 1 2

t

Figure 5.

If we consider the same caseT = 1/4 under the assumption of Theorem 2.3 whenq= 2,i.e.withλ > T /2

= 1/8, we can still apply (5.8) but only with T^∗= 3/4, and we obtain Eφ(0)≤ C

Z 3/4 0

Z (1/2)+λ (1/2)−λ

φ²t(x, t) dxdt

=C Z 1/4

0

Z (1/2)+λ (1/2)−λ

φ²_t(x, t) dxdt+C Z 1/2

1/4

Z (1/2)+λ (1/2)−λ

φ²_t(x, t) dxdt +C

Z 3/4 1/2

Z (1/2)+λ (1/2)−λ

φ²_t(x, t) dxdt

=C Z 1

0

a(t) Z

ω

φ²_t(x, t) dx+C Z 1/2

1/4

Z (1/2)+λ (1/2)−λ

φ²_t(x, t) dxdt, (5.11) but we need

Eφ(0)≤C Z 1

0

a(t) Z

ω

φ²_t(x, t) dxdt

=C Z 1/4

0

Z (1/2)+λ (1/2)−λ

φ²_t(x, t) dxdt+C Z 3/4

1/2

Z (1/2)+λ (1/2)−λ

φ²_t(x, t) dxdt. (5.12) to solve our problem. Note that the assumption “λ > 1/8” and this inequality correspond, as we can see in Figure 6, to the fact that each optic ray crosses the damping region during the period (0,1/4) or during the period (1/2,3/4).

0 x

1 2

t

Figure 6.

More generally, we prove the following new “on-off” observability inequalities:

Theorem 5.2. Let ` ∈N. Assume T = 1/(2`) and T /2< λ≤1/2. Then there existsC > 0such that, for allφsolution of (2.6),

Eφ(0) ≤ C

`−1

X

p=0

Z (2p+1)T 2pT

Z 1/2+λ 1/2−λ

φ²_t(x, t) dxdt=C Z 1

0

a(t) Z

ω

φ²_t(x, t) dxdt. (5.13)

Inequality (5.13) is exactly what we need to prove the exponential decay of the solutions of (1.2) whenq= 2, 1/T ∈2Nand 2λ > T.

Note that in the general case (q6= 2), the proof of Theorem 2.3 is based on Theorem 2.4. Theorem 2.4 is the generalization of Theorem 5.2 and its proof is similar to the proof of Theorem 5.2.

Remark. In fact repeating the arguments of the proof of Theorem 5.2, one shall prove that Eφ(0) ≤C

Z T^∗ 0

a(t) Z

ω

φ²_t(x, t) dxdt. (5.14)

withT^∗ :=KqT +ρat the place of q−1, where K∈N, ρ∈(0, T) and T^∗ >1−2λ. These are the optimal conditions: indeed, one cannot expect that (5.14) holds true withT^∗ < 1−2λ (this value is optimal when a(t) = 1 for allt), andT^∗ has necessarily to be writtenKqT +ρ: in the other case, equation (5.14) would be satisfied with a strictly smaller value ofK.

It is interesting to note that these results areoptimalin the following sense: ifT >2λ, then we can construct some initial conditions (φ⁰, φ¹) for whichEφ(0) = 1 while

`−1

X

p=0

Z (2p+1)T 2pT

Z 1/2+λ 1/2−λ

φ²_t(x, t) dxdt= Z 1

0

a(t) Z

ω

φ²_t(x, t) dxdt= 0. (5.15) This result correspond to the fact that, as we can see in Figure 7, there exists some optic rays that never cross the damping region at a time when the damping is effective. See Figure 7 in the case T = 1/4 with λ < T /2 = 1/8.

0 1 2

t x

Figure 7.

5.2.3. Other application of the observability inequalities: Exact controllability results

Theorem 5.2 also obviously gives new exact controllability results, applying the method H.U.M. of J.-L.

Lions [22]: given (u⁰, u¹)∈ H₀¹(Ω)×L²(Ω), there exists a controlh∈L²((∪^`p=0⁻¹(2pT,(2p+ 1)T), L²(ω)) such that the solution of the problem







utt−uxx=χω(x)h(t, x), x∈(0,1), t≥0, u(0, t) =u(1, t) = 0, t≥0,

(u(x,0), ut(x,0)) = (u⁰(x), u¹(x)), x∈(0,1),

(5.16)

satisfiesu(1−T) = 0 =ut(1−T). This implies that the controlhthat drives the system from the state (u⁰, u¹) to the rest acting only on the time intervals (2pT,(2p+ 1)T) for p= 0,· · ·, `−1, so, roughly speaking,“only half of the time”. We will generalize these results to prove that in fact we can find a control that drives the system from the state (u⁰, u¹) to the rest acting only on a finite number of well-chosen time intervals, whose total lenght is arbitrarily short(see [26]).

5.2.4. Open questions

1. It is an interesting open question to know what happens when 1/T ∈ 2N and 2λ= T. This should be compared to the results of pointwise stabilization or pointwise controllability (see, e.g.[14]).

2. Our proof does not allow us to obtain results for nonlinear on-off feedbacks.

3. It should also be interesting to study the problem withpositive negative feedbacks (1.4).

6. Proof of Proposition 5.1 (on-off boundary feedback when q = 2 )

We make the proofs in the casea0= 1. Our method can easily be extended to the general casesa0>0, in order to prove that the energy decays exponentially to zero.

6.1. The useful formula for the energy

Let us consider the absolutely continuous functionf : (−1,+∞)−→Rdefined on (−1,1) by:

∀y∈(−1,0), f(y) :=1 2

Z y 0

u⁰x(−z)−u¹(−z) dz,

∀y∈(0,1), f(y) := 1 2

Z y 0

u⁰_x(z) +u¹(z) dz, and on (1,+∞) by induction thanks to the expression

f⁰(y+ 1) =k(y)f⁰(y−1) withk(y) :=a(y)−1

a(y) + 1 a.e. y≥0. (6.1)

Note that this is equivalent to say that

f⁰(y+ 1) +f⁰(y−1) =−a(y)(f⁰(y+ 1)−f⁰(y−1)) a.e. y≥0. (6.2) Hence, using the d’Alembert’s formula, we easily verify that the following expression gives a solution of the problem (1.1) (and that this solution is unique):

∀(x, t)∈(0,1)×(0,∞), u(x, t) =f(t+x)−f(t−x).

(Relation (6.2) givesux(1, t) =−a(t)ut(1, t).) We see that the energy ofusatisfies

∀t≥0, Eu(t) = 1 2

Z 1 0

(u²_x(x, t) +u²_t(x, t)) dx= Z 1

−1

f⁰(t+s)²ds.

Now setN ∈Nandt∈(0,2) such that f⁰(t−1)∈R. We deduce from (6.1) that f⁰(t+ 2N+ 1) =k(t+ 2N)f⁰(t+ 2N−1) =· · ·=

YN m=0

k(t+ 2m)

!

f⁰(t−1).

Denote

KN(t) :=

YN m=0

|k(t+ 2m)|.

Hence we get that

Eu(2N+ 2) = Z 2

0

KN(t)²f⁰(t−1)²dt. (6.3)

Note that fort∈(0,2), the quantityf⁰(t−1) depends only on the initial conditions (u⁰, u¹). In the following, we will study the functionKN.

6.2. Proof of Proposition 5.1

Note that in this case the functionk satisfies:

(k(t) = 0 ift∈ G²:=S

n∈N[2nT,(2n+ 1)T), k(t) =−1 ift∈ B2:=S

n∈N[(2n+ 1)T,(2n+ 2)T).

Hence KN(t) = 1 if and only ift,t+ 2,· · ·,t+ 2N lie inB², andKN(t) = 0 in the other case. (G² is the set of the “good” values oft andB²is the set of the “bad” values of t).

Write

2 = 2`T+rT, with`∈Nand 0≤r <2. (6.4)

First assume that r = 0 (i.e. 1/T ∈ N): this implies that [T,2T) ⊂ [0,2)∩ B² and if t ∈ B², for example t∈[T,2T), thent+ 2 =t+ 2`T ∈[T + 2`T,2T+ 2`T)⊂ B2, and, in the same way,t+ 4,· · · ,t+ 2N are all inB2; henceKN(t) = 1 for allN ∈Nand allt∈ B2. Thus, for allN ∈N,

Eu(2N+ 2) = Z

[0,2)∩B2

KN(t)²f⁰(t−1)²dt+ Z

[0,2)∩G2

KN(t)²f⁰(t−1)²dt= Z

[0,2)∩B2

f⁰(t−1)²dt =: E_∞.

Consequently, since the energy is nonincreasing and its value at times 2N+ 2 remains constant, it is constant on [2,+∞) and equal to E_∞. This quantity depends only on the initial conditions (u⁰, u¹) and is non equal to zero if the initial conditions are well chosen. Note that for allt ≥2, Eu(t) =E_∞, andE_∞ = 0 if and only if the functiont7→f⁰(t−1) is equal to zero on [0,2)∩ B². In particular, ifu⁰ andu¹are such that the functionf is compactly supported in{t : t+ 1∈ B²∩[0,2)}, then the energy of the associated solution remains constant with time.

Now assume that 0< r≤1. (Note that in this case,dT =d(1/T,N) =r/2>0.) We claim thatKN(t) = 0 ifN ≥ ¹r. This follows from the following lemma (see the proof later):

Lemma 6.1. FixN ≥ ¹r. Givent∈[0,2), it is not possible fort,t+ 2,· · ·,t+ 2N to be all inB².

In other words, Lemma 6.1 means thateach optic ray touches the boundary point x= 1 in time at most2N and at an instant timet where the damping is effective. Note also that Lemma 6.1 implies thatKN(t) = 0 for allt∈[0,2), thus Eu(2N+ 2) = 0, which proves (5.1) in this case.

At last we assume that 1< r < 2. (Note that in this case, dT = 1−r/2 >0.) This case is completely analogous to the previous one. We claim that KN(t) = 0 ifN ≥ 2−¹r. This follows from the following lemma (see the proof later):

Lemma 6.2. FixN ≥ 2−¹r. Givent∈[0,2), it is not possible fort,t+ 2,· · ·,t+ 2N to be all inB².

On the same way, it implies KN(t) = 0 for all t ∈ [0,2) if N(2−r) ≥ 1, and soEu(2N + 2) = 0. This proves (5.1) in this case, which ends the proof of Proposition 5.1.

Proof of Lemma 6.1. We assume that 0< r ≤ 1 and we prove the result by contradiction: assume that t, t+ 2,· · ·,t+ 2N are all inB²for somet∈[0,2). Then firstt∈[0,2)∩ B²: hence there existsn0∈Nsuch that t∈[(2n0+ 1)T,(2n0+ 2)T); then using 2 = 2`T+rT with 0< rT ≤T, we deduce

t+ 2∈[(2n0+ 1)T+ 2`T,(2n0+ 2)T+ 2`T+T) = [(2`+ 2n0+ 1)T,(2`+ 2n0+ 2)T)

∪ [(2`+ 2n0+ 2)T,(2`+ 2n0+ 2)T+T).

Note that [(2`+ 2n0+ 1)T,(2`+ 2n0+ 2)T) ⊂ B² whereas [(2`+ 2n0+ 2)T,(2`+ 2n0+ 3)T) ⊂ G². Since t+ 2 ∈ B2, then t+ 2 ∈ [(2`+ 2n0+ 1)T,(2`+ 2n0+ 2)T); an easy induction argument shows that for all m∈ {0,· · ·, N}

t+ 2m∈[(2m`+ 2n0+ 1)T,(2m`+ 2n0+ 2)T).

Thus we deduce (withm=N and usingt≥(2n0+ 1)T),

(2n0+ 1)T + 2N ≤t+ 2N <(2N `+ 2n0+ 2)T, which implies that

rN T =N(2−2`T)< T, which is false.

Proof of Lemma 6.2. We assume that 1< r < 2 and we prove by contradiction that it is not possible fort, t+ 2,· · ·,t+ 2N to be all inB² ifN≥1/(2−r). Indeed, ift,t+ 2,· · ·,t+ 2N are all in B², then

t+ 2N ∈[(2N(`+ 1) + 2n0+ 1)T,(2N(`+ 1) + 2n0+ 2)T), hence (usingt≤(2n0+ 2)T),

(2n0+ 1)T + (2`+ 2)N T ≤t+ 2N <(2n0+ 2)T+ 2N, which implies that

N(2−r)<1.

Proof of the optimality of the result. Note that our result is optimal: assume that 0< r≤1 and denoteNT

the smallest integer such thatrNT ≥1. Hence we have:

1

r ≤NT < 1

r + 1−ε

for ε small enough. Then it is easy to verify that if t∈[T,(1 +ε)T)∩[0,2), then t, t+ 2, · · ·, t+ 2(N −1) are all inB² and so if the initial conditions are such that the corresponding functionf is supported in [T −1, (1 +ε)T−1)∩[−1,1), thenEu(2NT) =Eu(0), whereasEu(2NT + 2) = 0.

The same phenomenom occurs if 1≤r <2.

7. Proof of Theorem 2.1 (on-off boundary feedback in the general case)

Now we assume thata= 1 on [0, T),a= 0 on [T, qT) andaisqT-periodic. Define (G^q:=S

n∈N[qnT,(qn+ 1)T), B^q:=S

n∈N[(qn+ 1)T, q(n+ 1)T).

Write

2 =q`T +rT, with`∈Nand 0≤r < q. (7.1)

7.1. Stability if T is different of some exceptional values We prove the following:

Lemma 7.1. Assume that

r /∈ L:={r0= 0} ∪

rs,p= s

pq : p= 2,· · ·, q−1, s= 1,· · ·, p−1

· (7.2)

Then the energy of every solution achieves zero in finite time.

Note thatr∈ Lif and only if (2.1) is satisfied.

Proof of Lemma 7.1. First assume that there existsm≥1 andk∈Nsuch that

kq < mr≤kq+ 1 i.e. mr∈(0,1] modq. (7.3) Then we claim that the energy of every solution achieves zero in finite time: indeed denoteρ:=mr−kq∈(0,1]

and chooseN such that N ρ ≥ q−1; then if t, t+ 2, · · ·, t+ 2N mare all in B^q, then of courset, t+ 2m, t+ 4m,· · ·,t+ 2N mare all inB^q; but

t+ 2m=t+mqlT +mrT =t+ρT+q(ml+k)T =t+ρT modqT,

and thust, t+ρT,t+ 2ρT,· · ·, t+N ρT are all inB^q, which is not possible sinceρ∈(0,1] and N ρ≥q−1 (same proof as Lem. 6.1). HenceEu(2mN+ 2) = 0.

Let us specify the real numbers r that satisfy (7.3). Forr∈(0, q), denote ρ⁰ := ^r_q ∈(0,1). Condition (7.3) is equivalent to say that there exists m≥1 such that

{mρ⁰} ∈

0,1 q

, (7.4)

where {x}denotes the fractional part ofx. This is clearly true ifρ⁰ :=_q¹₀ withq⁰≥q(andm:= 1); this is also true ifρ⁰ := ^p_q⁰₀ withq⁰ ≥qand pgcd (p⁰, q⁰) = 1: indeed, there exists (p⁰⁰, q⁰⁰) such that p⁰⁰p⁰−q⁰⁰q⁰ = 1, and thenp⁰⁰ρ⁰=_q¹0 +q⁰⁰; at last (7.4) is also true ifρ⁰∈/Q, since in that case the set{{mρ⁰}, m∈N}is everywhere dense in [0,1). Thus if r /∈ L, equation (7.3) is satisfied, and the proof of Lemma 7.1 is complete.

7.2. Non stability if T is equal to one of those exceptional values We prove that ifr∈ L, then we have no more strong stability property.

First, for all q ≥2, the result is clear for the value r = r0 = 0. Indeed in this case we have 2 =q`T. It implies that: (t ∈ B^q) =⇒ (t+ 2∈ B^q). Thus it is sufficient to prove that there exists a non empty interval I⊂[0,2[∩B^q, which is clear sinceT <2.